a(n)=a+(n-1)d.
S(n)=na+n(n-1)d/2.
24 = a(2) = a + d,a = 24-d
S(13)=13a+13*6d > S(6)=6a+3*5d > S(14) = 14a + 7*13d,
13a + 13*6d > 6a + 15d,0 < 7a + 53d = 7(24-d) + 53d = 7*24 + 46d,d > -84/23.
6a + 15d > 14a + 7*13d,0 > 8a + 76d,0 > 2a + 19d = 2(24-d) + 19d = 48 + 17d,d < - 48/17.
-84/23 < d < -48/17.
S(n) = na + n(n-1)d/2 = n(24-d) + n(n-1)d/2 = (d/2)n^2 + n[24-3d/2]
= (d/2)[n^2 + n(48/d - 3) + (48/d - 3)^2/4 - (48/d - 3)^2/4]
= (d/2){[n+(48/d-3)/2]^2 - (48/d-3)^2/4}
-84/23 < d < -48/17,
-23/84 > 1/d > - 17/48,
-92/7 > 48/d > -17,
-92/7 - 3 > 48/d - 3 > -20.
-8 > -113/14 > (48/d-3)/2 > -10.
n-8 > n+(48/d-3)/2 > n-10.
S(8),S(9),S(10) 中的最大項(xiàng)為S(n)的最大項(xiàng).